Say we are in a tournament and are UTG plus 1. We want to steal the blinds and antes and raise with ATC because the table is playing tight. Say the table will only call a raise with top 5% hands.

How do we figure out the chance one player will have one? I'm not a math guy. Do we take 5 and multiply it by number of players left to act? So in this scenario it would be 40%. But that seems way too high. Thanks

My rough estimate is 36%.

The answer has to be lower than 40% since more than one villain can have a top 5% hand.

The answer has to be higher than 1 - .95^8 = 33.66%, which is what we get when we pretend there's no card removal.

Getting the exact answer would be tedious (plus it would require you to define the top 5% hands since there's more than one way to define that).

5%

For whatever reason, his question concerns UTG+1 and not UTG, so it's 1 - .95^7 ≈ 30.2% for the approximation.

If all one is interested in is if the chances are about 50%, then the following approximation will yield the geometric mean of your opponents calling frequency:

m ≈ 7/(10*n + 4)

where n is the number of active opponents left.

So for n = 7 remaining active opponents, you get:

m ≈ 7/(10*(7) + 4) ≈ 9.5%

This means that if your 7 remaining opponents are calling with their top 9.5% of their hands, you can make a pot sized bet with air and be profitable, since your pot odds on your bluff is 1 to 1. BTW, all this assumes you always lose if you get called, which is not actually true, so this is very conservative. According to Pokerstove, 9.5% looks something like 88+,ATs+,KTs+,QJs,AJo+,KQo.

You can verify this with 1 - .905^7 ≈ 50.3%, so it works quite well in this example.

It works fine with smaller n as well e.g. with n = 1:

m ≈ 7/(10*(1) + 4) = 50%

He multiplied by 8 so I figured he was talking about live 10-handed tables.

Nifty geometric mean formula. Its result is a little low due to the assumption of no card removal, but still very conservative for the reason you said.

Basically your formula says that 1 - nthRoot(.5) =~ 7/(10n+4)

And I see that it works for small and large n. Very cool!

50% of the time, they will either have it or they won't

^^ funny the first 1000 times it was posted.

nah, thats why its still funny

I have a simulation program that can do a problem like this, The chance that at least one of 7 players has a hand in the top 5% came out to be 31.4% for 5 million simulation trials. It was 35.1% for 8 players – pretty good guess Heehaw

may i ask what program?

35% or 36% ? I guess I should be shoving more often...

although the last 2 tournaments I was in, I managed to shove exactly when the BB had pocket aces. I prefer math to anecdotal evidence, it's easy to get gun-shy.

Is there an easy, quick way to figure out (roughly) the chances someone has a top 5 or 10% hand in each position? So UTG its 40, UTG plus one it's 35, UTG plus two it's 30, etc etc??

Excellent post! I wish I paid more attention in school though

well yea, 5/10% chance each person has a 5/10% hand??

It's about 1 - .95^n
where n is the number of players to go

Let # non-folded villains = n
Let p be the probability equivalent to the % you're interested in

(Edit #2 -- cliffs: take n*p - [small amount], and that small amount decreases with n. Example: for n=8 and p=.05 it's 5% because np gives 40% and Statman said the answer is 35%. For n=7 it's 3.5%. That's all I should have written, but I'll leave what I wrote below.)
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Let P(n,p) be the function approximating P(at least one villain has a p-hand)
Let x = your guess for the small amount of error

By (incomplete) inclusion-exclusion and ignoring card removal: P(n,p) = np - C(n,2)p^2 + x

= np[1-(n-1)p/2] + x

which gives us:
P(9, .05) = 36% + x
P(8, .05) = 33% + x
P(7, .05) = 29.75% + x
P(6, .05) = 26.25% + x
P(5, .05) = 22.5% + x
P(4, .05) = 18.5% + x
P(3, .05) = 14.25% + x
P(2, .05) = 9.75% + x

We see by Staman's results that as n decreases, so does x. (For n=8, x is about 2%. For n=7 it's about 1.5%.)

You also notice that OP's estimate of n*p becomes more accurate as n decreases, making my formula unnecessary for small n's.
For small n, just do P = np - y
where y is maybe 1% (and y decreases with n).

You could also do that for bigger n and just subtract a bigger y (e.g. for n=8 we see that y = 5%). That would be the quickest. And so actually, my formula is unnecessary period.

Edit: Not easy to do in your head though, compared to n*p.

It's called Hold'em Ranger - something I wrote in Excel using Visual Basic. It uses analytics and simulation to calculate opponent hand/range probabilities given opponent opening ranges and possible card removals through your hand and/or the board.

How long did that take to create? lol

if 5 people left behind you to act,
Probability of the person behind you who does not have 5% hand is (1259-your hand which is 1)/1326)probability of the next next person behind the person that behind you is (1258-1)/1326... so on up to 6 people, so you have a series of multiplication (1259*1258*1257*1256*1255*1254/(1326^6)=0.724 take 1 minus that number would be the probability of at least one of them have 5% of the hand which happen to be 27.6%. verify this answer with probability of 1 person who has hand, 2 person who has hand... up to 6 person who has top 5% of hand, it should add up to be 1.

You can simply use your method, 5% times ppl behind you for a quick and dirty approximation.

The very first version was 4 years ago and probably took me several months on very part time basis. Since then, I've been in the feature-creep mode where it does so many different things i have trouble remembering I can use it to sometimes answer 2p2 questions.

Nice approach but I think you have to make some adjustments.

First of all you’re assuming that hero does not have a 5% hand. That’s ok but not sure OP had that in mind. More important, you kept the denominator a constant 1326. If you’re going to assume hero has a card it should start off with 1325 and go down 1 for each player behind.

There are 0.95*1326 = 1260 non-5% hands (rounded up) . If we ignore what hero may have, the formula is 1- 1260/1326 * 1259/1325 * ….. When done for 6 players (your calculation) I got 26.43%, which should be exact with the 5% actually being 1260/1326 = 4.98%. My simulation gave me 26.65% using cards in top 4.99%.

Boy, how geeky can one get!!!

That definitely isn't exact. When a villain has one of the 1260 blank combos, that eliminates more than one blank combo for the next villain to have. It can also eliminate some of the 66 important combos e.g. if a villain has K2 that eliminates three of the KK's.

The denominators won't decrease by 1 either. They'll be 1326 followed by C(50,2) followed by C(48,2) and so on. The difference between 1326 and C(50,2) is 101. The next difference is 97.

Unfortunately, if you try fixing it by subtracting those differences from both the numerator and denominator, you get (for n=6) a bad answer of about 32%. That's because the numerators don't behave as predictably as the denominators.

Edit: furthermore, there can't be an exact solution in the form of 1-product. Your choice for the 1st numerator will have implications for the 2nd numerator and so on. So if going about it with that approach, you'd need to subtract multiple different products. However, this still might be better than the exact inclusion-exclusion solution (and I've little doubt it's better than addition), since I think PIE would be a pain for this problem. Now I'm curious to try this approach on the PIE problem I tackled (with the help of whosnext) in my recent Probability thread.

Of course. I was so focused on correcting the relatively minor errors made in the poster's method that I ignored the big error in the method itself.

As others have already pointed out, the answer to OP's query will depend upon several things.

First, what a "top 5%" hand means. So let's fix it to be some set of hands based upon some hand-ranking -- though we should be willing to admit the ultimate answer will depend slightly on what hand-ranking we choose.

Second, what do we assume for OP's hand? It is "cleaner" if we assume OP does not himself have a top 5% hand. But OP mentions an any-two-cards approach, meaning we could assume OP sometimes has a top 5% hand. So here is another "wrinkle" to take into account which would vary the answer.

Third, as heehaww and others point out, different non-top 5% rule out a differential number of top 5% hands. Using the hand-ranking based upon heads-up equity vs. a random hand, we see that 32 would not rule out any top-5 hands, 72 rules out a few (some 77 combos), while A2 rules out many more (some AA, AKs, AQs, AJs, AKo), etc..

Of course, combinatorics need to be obeyed if we try to do this exactly. As everyone knows, there are 6 combos of a pocket pair hand (AA), 4 combos of suited cards (AKs), and 12 combos of unsuited cards (AKo). Maybe to make things tractable, a reasonable approximation could be made as to the relationship between the top-5 and non-top-5 hand combos. I guess this "average overlap" factor would be of potential value in this and other questions.

Bottom line: there seems to be a continuum of answers to OP's query which balance accuracy with simplicity of solution. I wonder if anyone can derive a truly exact answer under a reasonable set of assumptions (a challenge to PIE I guess).